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Vaselesa [24]
3 years ago
14

When the volume of a gas is changed from 524 cm^3 to ____ cm^3, the temperature will change from 491 K to 297 K .

Chemistry
2 answers:
SpyIntel [72]3 years ago
5 0
V1/T1=V2/T2
524 cm^3/491 K =V2/297K

V2=(524 cm^3*297K)/491 K= 317K
Naily [24]3 years ago
4 0

Answer : The final volume of a gas is, 316.96cm^3

Explanation :

According to the Charles's law, the volume of a gas is directly proportional to the temperature of the gas at constant pressure and the number of moles of gas.

V\propto T

or,

\frac{V_1}{V_2}=\frac{T_1}{T_2}

where,

V_1 = initial volume of gas = 524cm^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 491 K

T_2 = final temperature of gas = 297 K

Now put all the given values in the above formula, we get the final volume of a gas.

\frac{524cm^3}{V_2}=\frac{491K}{297K}

V_2=316.96cm^3

Therefore, the final volume of a gas is, 316.96cm^3

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1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.
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Explanation:

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The formula for the calculation of moles is shown below:

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Thus,

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4 0
3 years ago
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
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Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

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For first isotope, 121 Sb :

% = x %

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For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

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<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

6 0
3 years ago
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